First, we must write the balanced equation for the decomposition of potassium chlorate:
#2KClO_3rarr2KCl+3O_2#
Note that the potassium chlorate and the oxygen are in the ratio #2:3#. That is to say, for every #2# moles of potassium chlorate which react, #3# moles of oxygen are produced.
The molar mass of potassium chlorate is #122.55"g/mol"#.
The formula for moles, #n#, is #n=m/M#, where #m# is the given mass and #M# the molar mass.
#n=283/122.55~~2.31"mol"# of potassium chlorate.
Remember that for every #2# moles of potassium chlorate which react, #3# moles of oxygen are produced.
So when #x# moles of oxygen are produced,
#2/3=2.31/x#
#2x=3*2.31#
#x~~3.47"mol"# of oxygen is produced.
You can rearrange the formula for moles to solve for #m#:
#n=m/M#
#m=nM#
The molar mass of oxygen is #32"g/mol"#, so we can input:
#m=32*3.47#
#m=110.88 \ "g"# of oxygen is produced.
